Potential energy of a system of particles is U=α3r3−β2r2, where r is distance between the particles. Here α and β are positive constants. Which of the following are correct for the system?
A
Equilibrium separation between the particles is αβ
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
For r=αβ, the equilibrium is unstable
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
For r=αβ, the equilibrium is neutral.
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
Work required to slowly move the particles to infinite separation from initial equilibrium position is βα
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is A Equilibrium separation between the particles is αβ Given: U=α3r3−β2r2
For equilibrium condition, F=0 ⇒−dUdr=0 ⇒−[−3α3r4−−2β2r3]=0 ⇒αr4−βr3=0 ⇒req=αβ
Now, d2Udr2=+4αr5−3βr4
At r=αβ, d2Udr2=4αβ5α5−3ββ4α4=β5α4
Since d2Udr2>0 ⟹stable equilibrium (as P.E is minimum)
Work required to slowly move the particle from r=αβ to ∞ =U∞−Ur=αβ =0−−β36α2 =β36α2